# Affine Algebraic Curves - Kunz - Exercise 1 - Chapter 1

#### Bernhard

I am reading Ernst Kunz book, "Introduction to Plane Algebraic Curves"

I need help with Exercise 1, Chapter 1 ...

Indeed ... I am a bit overwhelmed by this problem ..

Hope someone can help ... ...

To give a feel for the context and notation I am providing the start to Chapter 1, as follows:

Peter

#### johnsomeone

To make it conceptually clearer, imagine $$\displaystyle \mathbb{K} = \mathbb{C}$$ and $$\displaystyle \mathbb{K}_0 = \mathbb{R}$$.

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If $$\displaystyle f(x,y) \in \mathbb{R}$x, y$$$ has degree d which defines an algebraic curve $$\displaystyle \Gamma \subset \mathbb{C}^2$$ when considering solutions to $$\displaystyle f(x,y) = 0$$ in $$\displaystyle \mathbb{C}^2$$,

and $$\displaystyle y - mx - b \in \mathbb{R}$x, y$$$ is polynomial defining the line $$\displaystyle L \subset \mathbb{C}^2$$ when considering solutions to $$\displaystyle y - mx - b = 0$$ in $$\displaystyle \mathbb{C}^2$$,

then the points of $$\displaystyle (x,y) \in \mathbb{C}^2$$ which are on the intersection of $$\displaystyle \Gamma$$ and $$\displaystyle L$$ will have x-values satisfying $$\displaystyle f(x, mx+b) = 0$$ (and then corresponding y-values via $$\displaystyle y = mx - b$$).

Let $$\displaystyle p(x) = f(x, mx+b)$$. Then since $$\displaystyle m, b \in \mathbb{R}$$ and $$\displaystyle f(x,y) \in \mathbb{R}$x, y$$$, have that $$\displaystyle p(x) \in \mathbb{R}$x$$$.

Since $$\displaystyle f$$ has degree d, so does p.

If $$\displaystyle \{(x_k, y_k)\}_{k = 0}^d$$ are the d intersection points (including possibly multiplicity) of $$\displaystyle \Gamma$$ and $$\displaystyle L$$, have that $$\displaystyle p(x) = c \prod_{k = 1}^{d}(x - x_k) \in \mathbb{C}$x$$$.

So since $$\displaystyle p$$ has degree d, and is in $$\displaystyle \mathbb{R}$x$$$, if the first (d-1) of those $$\displaystyle x_k$$ are in $$\displaystyle \mathbb{R}$$, can you prove that the d-th one, the final one, $$\displaystyle x_d$$, is also in $$\displaystyle \mathbb{R}$$?

That's the point of this exercise.

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There's another possibility to consider: $$\displaystyle y - mx - b \in \mathbb{C}$x, y$$$ might not be capable of describing the line $$\displaystyle L$$, in which case you'll need: $$\displaystyle x - s \in \mathbb{C}$x, y$$$ to describe $$\displaystyle L$$.

But since the min poly of $$\displaystyle L$$ is in $$\displaystyle \mathbb{R}$x, y$$$, that tells you that $$\displaystyle s \in \mathbb{R}$$. In this case, points of intersection will satisfy $$\displaystyle f(s, y) = 0$$,

and so you'll be using a polynomial $$\displaystyle q(y) = f(s, y) \in \mathbb{R}$y$$$ to do the same reasoning as with above, except now using $$\displaystyle y$$'s and $$\displaystyle q(y)$$ instead of $$\displaystyle x$$'s and $$\displaystyle p(x)$$.

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The reasoning for general alg closed $$\displaystyle \mathbb{K}$$ with subfield $$\displaystyle \mathbb{K}_0$$ will work exactly the same way.

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#### johnsomeone

... will have x-values satisfying $$\displaystyle f(x, mx+b) = 0$$ (and then corresponding y-values via $$\displaystyle y = mx - b$$).
It obviously should be $$\displaystyle y = mx + b$$.

#### Bernhard

It obviously should be $$\displaystyle y = mx + b$$.
Thanks johnsomeone ... I really appreciate your help ...

I am just starting to work through you post in detail

Thanks again,

Peter